Let $\mathbf{m},$ $\mathbf{n},$ and $\mathbf{p}$ be unit vectors such that the angle between $\mathbf{m}$ and $\mathbf{n}$ is $\alpha,$ and the angle between $\mathbf{p}$ and $\mathbf{m} \times \mathbf{n}$ is also $\alpha.$  If $\mathbf{n} \cdot (\mathbf{p} \times \mathbf{m}) = \frac{1}{4},$ find the smallest possible value of $\alpha,$ in degrees.
Solution: By the scalar triple product,
\[\mathbf{p} \cdot (\mathbf{m} \times \mathbf{n}) = \mathbf{n} \cdot (\mathbf{p} \times \mathbf{m}) = \frac{1}{4}.\]Then
\[\|\mathbf{p}\| \|\mathbf{m} \times \mathbf{n}\| \cos \alpha = \frac{1}{4}.\]Also, $\|\mathbf{m} \times \mathbf{n}\| = \|\mathbf{m}\| \|\mathbf{n}\| \sin \alpha,$ so
\[\|\mathbf{p}\| \|\mathbf{m}\| \|\mathbf{n}\| \sin \alpha \cos \alpha = \frac{1}{4}.\]Since $\mathbf{m},$ $\mathbf{n},$ and $\mathbf{p}$ are unit vectors,
\[\sin \alpha \cos \alpha = \frac{1}{4}.\]Then $2 \sin \alpha \cos \alpha = \frac{1}{2},$ so
\[\sin 2 \alpha = \frac{1}{2}.\]The smallest possible angle that satisfies this is $\alpha = \boxed{30^\circ}.$